Tài liệu Physics exercises_solution: Chapter 35 docx

35.1: Measuring with a ruler from both S1 and S2to there different points in the antinodal line labeled m = 3, we find that the difference in path length is three times the wavelength o

35.1: Measuring with a ruler from both S1 and S2to there different points in the

antinodal line labeled m = 3, we find that the difference in path length is three times the

wavelength of the wave, as measured from one crest to the next on the diagram

35.2: a) At S1,r2  r1 4λ,and this path difference stays the same all along the

axis,

-y so m4.AtS2,r2 r1  λ, and the path difference below this point, along the negative y-axis, stays the same, so m4

b)

c) The maximum and minimum m-values are determined by the largest integer less

than or equal to .

λ

d

2

1

d so there will be a total of 15 antinodes between the sources (Another antinode cannot be squeezed in until the separation becomes six times the wavelength.)

35.3: a) For constructive interference the path diference is mλ,n0,1,2 The separation between sources is 5.00 m, so for points between the sources the largest possible path difference is 5.00 m Thus only the path difference of zero is possible This

occurs midway between the two sources, 2.50 m from A.

b) For destructive interference the path difference is (m 21)λ,m0,1,2

A path difference of  /λ 23.00 m is possible but a path difference as large as

00

9

2

/

λ  m is not possible For a point a distance x from A and 5.00x from B the path difference is

m 00 1 gives m 00 3 ) m

00

5

(

m 00 4 gives m 00 3 ) m

00

5

(

)

m

00

5

(





















x x

x

x x

x

x x

35.4: a) The path difference is 120 m, so for destructive interference:

120m λ 240m

2

b) The longest wavelength for constructive interference is λ120m

35.5: For constructive interference, we need r2 r1 mλ(9.00mx)xmλ

, 1 0, 1, 2, 3, For

m

8.25 m, 7.00 m, 5.75 m, 4.50 m, 3.25 m, 2.00 m, 75

0

) m 25 1 ( m 5 4 Hz) 10 2(120

) s m 10 00 3 ( m 5 4 2 m 5 4 2

λ m

5

8































m x

m

m f

mc m

x

3

,

2 

 (Don’t confuse this m with the unit meters, also represented by an “m”).

35.6: a) The brightest wavelengths are when constructive interference occurs:

nm

408 5

nm

2040

λ

and nm 510 4

nm 2040 λ

, nm 680 3

nm 2040 λ

λ





















s

m

d m

d

b) The path-length difference is the same, so the wavelengths are the same as part (a)

35.7: Destructive interference occurs for:

5 4

nm 2040 λ

and nm 3 58 5

3

nm 2040 λ

2

1





m

d

35.8: a) For the number of antinodes we have:

Hz) 10 (1.079 m) (12.0

) s m 10 (3.00 λ

8















df

mc d

m

the maximum integer value is four The angles are 13.4, 27.6, 44.0, and 67.9 for m0,1, 2, 3, 4

b) The nodes are given by sin  ( 12)λ 0.2317(m12)

d

m

3 , 2 , 1 , 0 for 2 54 , 4 35 , 3

20

,

65

m 2.20

m) 10 (2.82 m) 10 (4.60 λ





R

y d d

R

y

35.10: For bright fringes:

mm 1.14 m 10 1.14 m

0.0106

m) 10 (5.02 (20) m) (1.20











m y

Rm

d

35.11: Recall

m 10 4.50

m) 10 (5.00 ) m (0.750 )

2 3 ( λ

4

7 2

3























d

Rλ y y y d

Rm

y m

8.33 10 4 m 0.833mm



35.12: The width of a bright fringe can be defined to be the distance between its two

adjacent destructive minima Assuming the small angle formula for destructive

interference

, λ ) ( 21

d

m R

 the distance between any two successive minima is

mm

8.00 m)

10 (0.200

m) 10 (400 m) 00 4 (

λ

3

9











d R y

y n n

Thus, the answer to both part (a) and part (b) is that the width is 8.00 mm.

35.13: Use the information given about the bright fringe to find the distance d between

m 10 4.84

m) 10 (600 m) (3.00 λ

so ), 6 35 Eq (

3 9 1

1 1



















y

R d d

R y (R is much greater than d, so Eq.35.6 is valid.)

The dark fringes are located by dsin (m21)λ,m0,1, 2, The first order dark fringe is located by sin λ2 2d, whereλ2 is the wavelength we are seeking

d

R R

R

y

2

λ sin

We want λ such that 2 y  y1 This gives andλ λ 1200

2

λ λ

1 2 2

d

R d

R

nm

35.14: Using Eq.35.6 for small angles,

,

λ

d

m R

y m 

we see that the distance between corresponding bright fringes is

mm 3.17 m)

(10 470) (660 m) 10 (0.300

(1) m) (5.00











d Rm

y

35.15: We need to find the positions of the first and second dark lines:

m 0.0541 )

tan(8.79 m)

0.350 ( tan

79 8 m 10 2(1.80

m 10 5.50 arcsin

2

λ arcsin

1 1

6

7 1

















































R

y

d





























m) 10 2(1.80

m) 10 3(5.50 arcsin

2

λ

7 2

d



 y2Rtanθ2 (0.350m)tan(27.30)0.1805m

The fringe separation is then y y2 y1 0.1805m0.0541m0.1264m

35.16: (a) Dark fringe implies destructive interference.

λ 2

1 sin 

d

m 10 64 1 0 11 sin 2

m 10 624 sin

2



















d

(b) Bright fringes: d sinθmax mmaxλ

The largest that θ can be is 90 ,som d/λ 1.646241010 mm 2.6Sincem

6











maximum value is 2 There are 5 bright fringes, the central spot and 2 on each side of it Dark fringes: d sinθ  m  21λ This equation has solutions for θ11.0;34.9; and

6

72 

 Therefore, there are 6 dark fringes

35.17: Bright fringes for wavelength λare located by d sin mλ First-order (m1)

is closest to the central bright line, so sin λ/d

































0.401 and

m) 10 m)/(0.100 10

(700 sin

gives

nm

700

λ

0.229 and

m) 10 m)/(0.100 10

(400 sin

gives

nm

400

λ

3 9

3 9









The angular width of the visible spectrum is thus 0.4010.2290.172

m 10 4.20

m) 10 (4.50 m) 1.80 ( λ

3

7





















y

R d d

R

y

35.19: The phase difference  is given by  (2d/λ)sin(Eq.35.13)

rad 1670 0

23 sin ] m) 10 (500 m) 10 340

0

(

2



λ

difference Path







radians 119

cm 2

cm 486 cm 524











0

2

0cos ( 2) I (cos30.0 ) 0.750I I

b) 60.0(/3)rad

Eq.(35.11): (2 /λ)(r2 r1),so

( /3)/2 λ λ/6 80nm λ

) 2 /

(

)

35.22: a) The source separation is 9.00 m, and the wavelength of the wave is

m

0 20 Hz 10 50

1

m/s 10 00

3









f

c

So there is only one antinode between the sources ),

0

(m and it is a perpendicular bisector of the line connecting the sources

b)      sin

m) (20.0

m) 00 9 ( cos sin

cos 2

0

2 0

2

λ

d I

I I

cos2 ((1.41)sin )

I



; 295 0 45

580 0 30

for

So, θ  , I  I0;θ , I  I0

θ 60, I 0.117 I0;θ90, I 0.026I0

35.23: a) The distance from the central maximum to the first minimum is half the

distance to the first maximum, so:

m

10 88 8 m)

10 2(2.60

m) 10 (6.60 m) 700 0 ( 2

4

7



















d

R y

b) The intensity is half that of the maximum intensity when you are halfway to the first minimum, which is 4.4410 4m.Remember, all angles are small

35.24: a) 2.50m,

Hz 10 20 1

m/s 10 00 3









f

c

and we have:

rad

4.52 m)

8 1 ( m 50 2

2 ) ( λ

2

2

2

rad 52 4 cos 2

0

2

I

























m 10 1.30

m) 10 (5.50 m) 900 0 (

4

7

















d

R y

So the distance to the first minimum is one half this, 1.91 mm

b) The first maximum and minimum are where the waves have phase differences of zero and pi, respectively Halfway between these points, the phase difference between the waves is So:

2



W/m 10

00 2 2 4

cos 2

0

2 0



































35.26: From Eq (35.14), sin

λ

cos2











I So the intensity goes to zero when the

cosine’s argument becomes an odd integer of   sin ( 1/2) 

λ : is That

d

), 2 / 1 (

λ

sin  m

35.27: By placing the paper between the pieces of glass, the space forms a cavity whose

height varies along the length If twicethe height at any given point is one wavelength (recall it has to make a return trip), constructive interference occurs The distance

between the maxima (i.e., the # of meters per fringe) will be

0.0235 rad

10 095 4 m) ) 1500 / 1 ((

2

m 10 5.46 arctan 2

λ arctan tan

2

λ 2

































x h

l



35.28: The distance between maxima is

cm

0369 0 m)

10 2(8.00

cm) (9.00 m) 10 56 6 ( 2

λ

5

7











h

l x

So the number of fringes per centimeter is 1  27.1

35.29: Both parts of the light undergo half-cycle phase shifts when they reflect, so for

) 42 1 ( 4

m 10 50 6 4

λ 4

n t

35.30: There is a half-cycle phase shift at both interfaces, so for destructive

interference:

nm

5 80 4(1.49)

nm 480 4

λ 4



n t

35.31: Destructive interference for λ1 800nm incident light Let nbe the refractive index of the oil There is a λ/2 phase shift for the reflection at the air-oil interface but no phase shift for the reflection at the oil-water interface Therefore, there is a net λ/2

phase difference due to the reflections, and the condition for destructive interference is

)

/

λ

(

2t m n Smallest nonzero thickness means m1,so2tnλ1

The condition for constructive interference with incident wavelength λis

on so and nm, 320 λ

2,

for

nm 533 λ

,

1

for

nm 1600 λ

,

0

for

nm

800 λ

where ), /(

λ λ so ,

λ

2

But

λ ) ( 2 and ) / λ

)(

(

2

1 2

1 1

1

2

1 2

1





























m

m

m

m tn

m tn n

m

t

The visible wavelength for which there is constructive interference is 533 nm

35.32: a) The number of wavelengths is given by the total extra distance traveled,

divided by the wavelength, so the number is

5 36 m

10 6.48

(1.35) m) 10 76 8 ( 2 λ

2

6 0









x

b) The phase difference for the two parts of the light is zero because the path

difference is a half-integer multiple of the wavelength and the top surface reflection has a half-cycle phase shift, while the bottom surface does not

35.33: Both rays, the one reflected from the pit and the one reflected from the flat

region between the pits, undergo the same phase change due to reflection The condition for destructive interference is 2t (m 21)(λ/n), where n is the refractive index of the plastic substrate The minimum thickness is for m0, and equals

m

0.11 nm 110 8)]

nm/[(4)(1

790

)

4

/(

t

35.34: A half-cycle phase change occurs, so for destructive interference

nm

180 2(1.33)

nm 480 2

λ

2



n

t

35.35: a) To have a strong reflection, constructive interference is desired One part of

the light undergoes a half-cycle phase shift, so:

2 1

nm 771 2

1

(1.33) nm) 290 ( 2 2 1

2 λ

λ 2

1 2











 













 













 















 



m m

m

dn n

m

value of zero, the wavelength is not visible (infrared) but for m1, the wavelength is

514 nm, which is green

b) When the wall thickness is 340 nm, the first visible constructive interference occurs again for











 





2 1

nm 904 λ yields and 1

m

35.36: a) Since there is a half-cycle phase shift at just one of the interfaces, the

minimum thickness for constructive interference is:

nm

3 74 4(1.85)

nm 550 4

λ 4



n t

b) The next smallest thickness for constructive interference is with another half

nm

223 )

85 1 ( 4

nm 550 3 4

λ 4









n t

2

) m 10 33 6 ( 1800 2











x

2

m) 10 06 6 ( 818 2

1















2

m) 10 02 5 ( 818 2

2















b) The net displacement of the mirror is the difference of the above values:

mm

0.043 mm

0.205 mm

248 0 2





35.39: Immersion in water just changes the wavelength of the light from Exercise

1.33

mm 833 0



n

y dn

R

35.11

35.40: Destructive interference occurs 1.7 m from the centerline.

2 2

1  (12.0m) (6.2m)

2 2

2  (12.0m) (2.8m)

For destructive interference, r1 r2 λ/21.19mand λ2.4m.The wavelength we have calculated is the distance between the wave crests

Note: The distance of the person from the gaps is not large compared to the separation of

the gaps, so the path length is not accurately given by d sin

35.41: a) Hearing minimum intensity sound means that the path lengths from the

individual speakers to you differ by a half-cycle, and are hence out of phase by 180 at  that position

b) By moving the speakers toward you by 0.398 m, a maximum is heard, which means that you moved the speakers one-half wavelength from the min and the signals are back

in phase Therefore the wavelength of the signals is 0.796 m, and the frequency is

m 0.796

m/s

340

λ 

 v

c) To reach the next maximum, one must move an additional distance of one

wavelength, a distance of 0.796 m

35.42: To find destructive interference, λ

2

1 )

200

1









 











d

λ 2

1 2

1 λ 2 1

m 000 , 20

λ 2

1 2

λ 2

1 )

m 200

(

2

2 2

2 2











 













 















 

























 









m m

x

m x m

x x

The wavelength is calculated by 51.7m

Hz 10 80 5

s m 10 00 3

8











f c

m 0 20

; 3 and m, 1 90 : 2 and m, 219 :

1 and m, 761 :



35.43: At points on the same side of the centerline as point A the path from , B is longer than the path from A and the path difference d sin θ puts speaker A ahead of , speaker B in phase Constructive interference occurs when

  0.2381, 0,1,2,

3

2 λ

3

2

sin

, 2 , 1 , 0 , λ 2

1 6

λ

sin













 













 















 





m m

d m

m m

d



, 4

; 8 60 , 3

; 4 39 , 2

; 4 23 , 1

; 13

.

9

,

At points on the other side of the centerline, the path from A is longer than the path from

B , and the path difference d sin θ puts speaker A behind speaker B in phase

Constructive interference occurs when

  0.2381, 0,1,2,

3

1 λ

3

2

sin

, 2 , 1 , 0 , λ 2

1 6

λ

sin













 













 















 





m m

d m

m m

d





, 4

; 5 52 , 3

; 7 33 , 2

; 5 18 , 1

;

55

4

,

35.44: First find out what fraction the 0.159 ms time lag is of the period.

) Hz (1570 s) 10 159 0 ( ) s 10 159 0 ( 10 159



















T

s t

, 250 0



t so the speakers are 1 period out of phase Let A be ahead of B in 4 phase

m 210 0 Hz 1570

s m 330

centerline of

side

s

A

On ' : Since A is ahead by 41 period, the path difference must retard B’s phase enough so the waves are in phase











































6 60 m

422 0

m 210 0 4

7 sin

9 21 m

0.422

m 0.210 4

3 sin

,

λ 4

7 , λ 4

3 sin

2 2

1 1











d

On ' B s side of centerline: The path difference must now retard A’s sound by



,

λ

,

λ 45

4

1

  λ, gives7.2,38.5

4

5 , λ 4

1

d

35.45: a) If the two sources are out of phase by one half-cycle, we must add an extra

half a wavelength to the path difference equations Eq (35.1) and Eq (35.2)

This exactly changes one for the other, for and 21 ,

2

m

in any integer

b) If one source leads the other by a phase angle , the fraction of a cycle difference is

2



Thus the path length difference for the two sources must be adjusted for both destructive and constructive interference, by this amount So for constructive

inference:r1r2 (m 2)λ, and for destructive interference,

λ ) 2 2 1

(

2

1r  m  

r